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Answer 1 · 2017-12-15T18:34:15

#1/2arcsin((2x)/3)+c#

#int1/sqrt(9-4x^2)dx#

it can be rewritten

#int1/sqrt(3^2-(2x)^2)dx#

trigonometrical substitution

#2x=3sin(theta)#

#dx=(3cos(theta)d(theta))/2#

#sqrt(9-4x^2)=3cos(theta)#

in the integral

#int1/sqrt(3^2-(2x)^2)dx=int(3cos(theta)d(theta))/(2*3cos(theta)#

#int(3cos(theta)d(theta))/(2*3cos(theta)#=#int(d(theta))/2#

#1/2int(d(theta))=1/2(theta)+c#

but
#theta=arcsin((2x)/3)#

#1/2int(d(theta))=1/2arcsin((2x)/3+c)#

#int1/sqrt(9-4x^2)dx=1/2arcsin((2x)/3)+c#